Solution:

We know that the normalized frequency is given by,

$V=\frac{2 \pi a}{\lambda}\left(n_1^2-n_2^2\right)^{1 / 2}$

Also,

$ N A=\sqrt{n_1^2-n_2^2} \quad \therefore V=\frac{2 \pi a}{\lambda} N A\\ $

Given

$V=26.6, a=25 \mu \mathrm{m}, \lambda=1300 \mathrm{~nm}=1.3 \mu \mathrm{m}$

$ N A=\frac{V \lambda}{2 \pi a}\\ $

$ =\frac{26.6 \times 1.3}{2 \pi \times 25} $

$ =0.22 $